Aball of mass m is released at the top edge of a frictionless half-pipe with a radius of curvature r. at the bottom of the pipe, the ball is moving with a speed v. what is the normal force acting on the ball at this point?

Explanation:

At the bottom of the pipe :

= Normal force acting in upward direction

 = mass of the ball

 = weight of the ball acting in downward direction = mg

 = speed of the ball at the bottom

 = radius of curvature

Force equation for the motion of ball is given as

the current is 0.6

voltage ÷ resistance = current

so 12 ÷ 20 = 0.6

54 degrees incident angle = reflected angle

angle is from normal turned to line of ray